The given equation can be written as: \[ \tan^{-1} \sqrt{x^2 + x} = \frac{\pi}{2} - \sin^{-1} \sqrt{x^2 + x + 1}. \] Simplify using the trigonometric identity \( \sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x \): \[ \tan^{-1} \sqrt{x^2 + x} = \cos^{-1} \sqrt{x^2 + x + 1}. \] Using the property \( \tan^{-1} x = \cos^{-1} \left(\frac{1}{\sqrt{1+x^2}}\right) \), we rewrite: \[ \cos^{-1} \frac{1}{\sqrt{x^2 + x + 1}} = \cos^{-1} \sqrt{x^2 + x + 1}. \] Equating the arguments (as the \( \cos^{-1} \) function is one-to-one for \( [0, \pi] \)): \[ \frac{1}{\sqrt{x^2 + x + 1}} = \sqrt{x^2 + x + 1}. \] Squaring both sides: \[ x^2 + x + 1 = 1. \] Simplify: \[ x^2 + x = 0 \quad \Rightarrow \quad x(x + 1) = 0. \] Thus, \[ x = 0 \quad \text{or} \quad x = -1. \]